In a uniform electric field, a cube of side 1 | vedclass

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Question

Energy $=\frac{1}{2} \in_{0} \mathrm{E}^{2}$ (volume)

$8.85 	imes 10^{-6}=\frac{1}{2} 	imes 8.85 	imes 10^{-12} \mathrm{E}^{2}\left(10^{-6}igh

Vedclass · Accepted Answer

$100\sqrt 2 \,\ \frac{V}{m}$

Vedclass · Answer

Energy $=\frac{1}{2} \in_{0} \mathrm{E}^{2}$ (volume)

$8.85 	imes 10^{-6}=\frac{1}{2} 	imes 8.85 	imes 10^{-12} \mathrm{E}^{2}\left(10^{-6}ight)$

$\mathrm{E}=\sqrt{2} 	imes 10^{6} \mathrm{\,V} / \mathrm{m}$

flux $(\phi)=E A$

$=\sqrt{2} 	imes 10^{+6} 	imes 10^{-4}=100 \sqrt{2}\,(\mathrm{V}/ \mathrm{m})$

In a uniform electric field, a cube of side $1\ cm$ is placed. The total energy stored in the cube is $8.85\ \mu J.$ The electric field is parallel to four of the faces of the cube. The electric flux through any one of the remaining two faces is.

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